3

I am very confused with the concept of projection with the introduction of immersion and submersion. By local immersion/submersion theorem, for a simmersion/submersion $f$, there is is a canonical immersion/submersion locally is equal to $f$.

So does this imply that locally, for a immersion/submersion of $f$ at $x$, it is equal to the projection and inverse projection function?

Are these three all open maps?

Thank you~

WishingFish
  • 2,472

1 Answers1

8

Well, projections are open since basic open sets in the product topology are $U \times V$ with $U$ and $V$ open, so projecting down leaves you with $U$ and you're done.

Now, since submersion is locally just a projection, it follows that it is also open.

Immersions need not be open though. For example, take canonical immersion $\mathbb R^1 \to \mathbb R^2$. The domain is surely open but the image is just a line which can't be open in the plane.

Marek
  • 6,346
  • that's amazing, your explanation is so clear, thanks Marek. Though I am not sure with the proof of the projection is open. Your proof makes sense to me, but I was baffled by the complex construction in this proof: http://math.stackexchange.com/questions/247542/projection-is-an-open-map. Is it necessary? Or it is just a more rigorous version? Thanks!!!!! – WishingFish Jul 05 '13 at 06:35
  • Yes, I skipped the details. Full rigorous proof should look like that. – Marek Jul 05 '13 at 06:43
  • And you're welcome :) – Marek Jul 05 '13 at 06:43